Video Transcript
1) Use Euler’s formula to express
𝑒 to the negative 𝑖𝜃 in terms of sine and cosine. 2) Given that 𝑒 to the 𝑖𝜃 times
𝑒 to the negative 𝑖𝜃 equals one, what trigonometric identity can be derived by
expanding the exponential in terms of trigonometric functions?
For part one, we’ll begin by
rewriting 𝑒 to the negative 𝑖𝜃. It’s the same as 𝑒 to the 𝑖 times
negative 𝜃. We can now apply Euler’s
formula. Since Euler’s formula says that 𝑒
to the 𝑖𝜃 is equal to cos 𝜃 plus 𝑖 sin 𝜃, we can see that 𝑒 to the 𝑖 negative
𝜃 is equal to cos of negative 𝜃 plus 𝑖 sin of negative 𝜃. And then, we recall the properties
of the cosine and sine functions. Cos is an even function. So cos of negative 𝜃 is equal to
cos of 𝜃. Sin, however, is an odd function. So sin of negative 𝜃 is the same
as negative sin 𝜃. And we can therefore rewrite our
expression. And we see that 𝑒 to the negative
𝑖𝜃 is the same as cos 𝜃 minus 𝑖 sin 𝜃.
Now, let’s consider part two of
this question. We’re going to use the answer we
got from part one. When we do, we can see that 𝑒 to
the 𝑖𝜃 times 𝑒 to the negative 𝑖𝜃 is the same as cos 𝜃 plus 𝑖 sin 𝜃 times
cos 𝜃 minus 𝑖 sin 𝜃. Let’s distribute these parentheses, perhaps noticing that this is an expression factored using the difference of two squares. cos 𝜃 times cos 𝜃 is cos squared 𝜃. cos 𝜃 times negative 𝑖 sin 𝜃 is negative 𝑖 cos 𝜃 sin 𝜃. We then get plus 𝑖 sin 𝜃 cos 𝜃
and 𝑖 sin 𝜃 times negative 𝑖 sin 𝜃 is negative 𝑖 squared sin squared 𝜃. We see then negative 𝑖 cos 𝜃 sin
𝜃 plus 𝑖 cos 𝜃 sin 𝜃 is zero. And of course, we know that 𝑖
squared is equal to negative one.
So we can simplify this
somewhat. And we see that 𝑒 to the 𝑖𝜃
times 𝑒 to the negative 𝑖𝜃 is cos squared 𝜃 plus sin squared 𝜃. We were told however that 𝑒 to the
𝑖𝜃 times 𝑒 to the negative 𝑖𝜃 was equal to one. So you can see that we’ve derived
the formula sin squared 𝜃 plus cos squared 𝜃 equals one. This is a fairly succinct
derivation of the trigonometric identity sin squared 𝜃 plus cos squared 𝜃 equals
one.
